Question

On a normally distributed anxiety test with mean 48 and standard deviation 4, approximately what anxiety test score would put someone in the top 5 percent?

Answer 1

Answer:

54.56

Step-by-step explanation:

Given:

Mean, u = 48

Standard deviation, $\sigma$ = 4

Test score = 5%

Required:

Find the raw score, X.

Having a test score of 5% means the Zscore has to correspond to 95%, ie, 1 - 0.05 = 0.95.

Thus, $Z_0_._9_5 = 1.64$

To get X, use the formula:

$mu + z * \sigma$

= 48 + (4 * 1.64)

= 48 + 6.56

= 54.56

The anxiety test score that would put someone in the top 5% is 54.56